e^(i*pi) = -1: pi = 0 ?

Date: 10/17/97 at 11:29:15
From: John K. Koehler
Subject: Weird tricks
Dr. Math,
I know that from a certain trig identity we get the equation
e^(i*pi) = -1. I have been playing around with this equation and have
found some disturbing things. I hope that you can help me.
First:
e^(2*i*pi) = 1
e^(-2*pi) = 1 (raised both sides to i and 1^n is 1)
-2*pi = 0 (took the ln of both sides and ln(1) = 0)
pi = 0 !
Second:
i*pi = ln(-1) {1}
ln(-n) = ln(-1) + ln(n)
therefore by {1}
ln(-n) = i*pi + ln(n)
Finally:
i^2 = -1
e^(i*pi) = i^2
e^(-pi) = i^(2i)
e^(-pi/2) = i^i
e^(-pi/2) is a real number and therefore so is i^i
Can you please tell me why I get such ridiculus results?
I asked my calc2 teacher last year - but all he could tell me
was that imaginary numbers don't work like reals, and I still
don't see how that would change anything.
Thanks in advance,
John

Date: 10/17/97 at 13:08:36
From: Doctor Bombelli
Subject: Re: Weird tricks
Well, John, not all the results you have are ridiculous! In fact, you
and your teacher are both right; complex numbers don't always work
like real numbers, but sometimes they do.
Complex numbers can be written in many equivalent ways:
z = x+iy
z = re^*(it) where t is the angle the point (x,y) makes with the
positive x axis.
t is called the argument of z--arg(z)
z = r[cos(t)+i sin(t)]
In fact, the definition of e^(it) is cos(t)+i sin(t).
This is why we have e^(-ipi) = cos(pi)+i sin(pi)= -1.
We would like w = log(z) and z = e^w to match up, just as for real
numbers.
Let z = re^(it) and w = u+iv.
z = e^w = e^(u+iv) = e^u*e^(iv) and z is also re(^it)
Match the real part and the "i part" (the imaginary part).
r = e^u and e^(it) = e^(iv)
So u = ln(r) and v can be any value with cos(v) = cos(t)
[since e^(it) is cos(t)+i sin(t)].
So we say, for complex numbers, ln(z)=ln(r)+i arg(z)+i2kpi
(k an integer). In fact, ln(z) has many values.
We also define, in this way, that
z^a = e^[a ln(z)] when a is complex.
(Check that this works for real numbers a, also.)
Note that 1^a = e^[ln(1)] = e^[0+i2kpi], so only one of the answers
is e^0 = 1.
Here is your problem:
e^(2*i*pi) = 1
e^(-2*pi) = 1 (raised both sides to i and 1^n is 1)
-2*pi = 0 (took the ln of both sides and ln(1) = 0)
pi = 0
In steps 2 and 3 you don't have one-to-one functions any more.
(The reason e^x = 1 implies x = 0 is because the real logarithm
function is one-to-one. It is kind of like why x^3 = 1 gives x = 1
and x^2 = 1 gives x = 1 or -1. The cube root is one-to-one, but
the square root isn't.)
Now in your second experiment:
i*pi = ln(-1) {1}
ln(-n) = ln(-1) + ln(n)
therefore by {1}
ln(-n) = i*pi + ln(n)
This is okay.
ln(-n) = ln(n) + i pi + i 2kpi from the definition above.
In your third experiment:
i^2 = -1
e^(i*pi) = i^2
e^(-pi) = i^(2i)
e^(-pi/2) = i^i
e^(-pi/2) is a real number and therefore so is i^i
i^i is e^[i ln(i)], and ln(i) is ln(1) + i pi/2 + i 2kpi
So... i^i = e^[0 + i^2 pi/2+ i^2 2kpi] = e^[- pi/2 - 2kpi]
which is a real number.
I commend you on playing around with this stuff. That is how new
mathematics is learned, on all levels.
-Doctor Bombelli, The Math Forum
Check out our web site! http://mathforum.org/dr.math/